A uniform rod of length 2 m and of mass 34 kg is suspended by two vertical strings - Leaving Cert Applied Maths - Question 7 - 2009
Question 7
A uniform rod of length 2 m and of mass 34 kg is suspended by two vertical strings.
One string is attached to a point 20 cm from one end and can just support a mass... show full transcript
Worked Solution & Example Answer:A uniform rod of length 2 m and of mass 34 kg is suspended by two vertical strings - Leaving Cert Applied Maths - Question 7 - 2009
Step 1
Find the length of the section ab of the rod within which the 3.4 kg mass can be attached without breaking either string.
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Answer
To find the length of the section ab:
Consider the moments about point c:
The weight of the 3.4 kg mass will create a moment about point c. The equation for moments can be written as:
F1(1.5)=3.48g(34)(0.8)
To ensure balance, the upward moments must equal the downward moments:
20.74g(1.5−x)=3.48g(1+34)(0.8)
Solving this gives:
x=1.15extm
Consider the moments about point d:
Similarly, treating moments about point d:
F1(1.5)=3.48g(34)(0.7)
Which leads to:
17g(1.5−x)=3.48g(34)(0.7)
Thus, solving results in:
x=1extm
Determine the length of ab:
By considering distances, we find:
∣ab∣=0.15extmor15extcm
Step 2
Find, in terms of W, the reaction at b.
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Answer
To find the reaction at point b:
Establish the moment about point a:
R(2pextcos20)=Wextp(cos0)
Where R represents the reaction at point b. From this, solve for R:
R=2W
Considering horizontal and vertical components:
For horizontal: Rextsinheta=Wextsin0
For vertical: Rextcosheta+Wextsin20=W
Rearranging provides:
2W cos 0=Wextso
extcos20extcos20=2extcos2heta
Show that cos θ = 2 cos 2θ:
Reorganize to conclude:
cos 20extcos20=2extcos2θ
Hence the derivation verifies the relationship.
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